# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.31

• Last Updated : 07 Apr, 2021

### Question 1. ∫(x2 + 1)/(x4 + x2 + 1)dx

Solution:

We have,

∫(x2 + 1)/(x4 + x2 + 1)dx

= ∫x2(1 + 1/x2)/x2(x2 + 1 + 1/x2)dx

= ∫(1 + 1/x2)/(x2 + 1 + 1/x2)dx

= ∫(1 + 1/x2)/(x2 + 1/x2 − 2 + 2 + 1)dx

= ∫(1 + 1/x2)/[(x − 1/x)2 + 3]dx

Let x − 1/x = t.

Differentiating both sides, we get,

(1 + 1/x2)dx = dt

So, our equation becomes,

= ∫1/(t2 + 3)dt

= (1/√3) tan-1(t/√3) + c

= (1/√3) tan-1[(x2 − 1)/√3x] + c

### Question 2. ∫√(cotθ)dθ

Solution:

We have,

∫√(cotθ)dθ

Let cotθ = x2

Differentiating both sides, we get,

=> −cosec2θdθ = 2xdx

=> dθ = −2xdx/cosec2θ

=> dθ = −2xdx/(1 + cot2θ)

=> dθ = −2xdx/(1 + x4)

So, our equation becomes,

= −∫2x2/(1 + x4)]dx

= −∫2/(x2 + 1/x2)dx

= −∫(1 + 1/x2 + 1 − 1/x2)/(x2 + 1/x2 + 2 − 2)dx

= −∫(1 + 1/x2)/[(x − 1/x)2 + 2]dx − ∫(1 − 1/x2)/[(x + 1/x)2 − 2]dx

Let x − 1/x = t, so we get, (1 + 1/x2)dx = dt

Let x + 1/x = y, so we get, (1 − 1/x2)dx = dy

So, our equation becomes,

= −∫dt/(t2 + 2) − ∫dy/(y2 − 2)

= −(1/√2) tan-1[t/√2] − (1/2√2) log |(y − √2)/(y + √2)| + c

= −(1/√2) tan-1[(x2 − 1)/√2x] − (1/2√2) log |(x2 + 1 − √2x)/(x2 + 1 + √2x)| + c

= −(1/√2) tan-1[(cotθ − 1)/√(2cotθ)] − (1/2√2) log |(cotθ + 1 − √(2cotθ))/(cotθ + 1 + √(2cotθ))| + c

### Question 3. ∫(x2 + 9)/(x4 + 81)dx

Solution:

We have,

∫(x2 + 9)/(x4 + 81)dx

= ∫x2(1 + 9/x2)/x2(x2 + 81/x2)dx

= ∫(1 + 9/x2)/[(x − 9/x)2 + 18]dx

Let x−9/x = t, so we have, (1 + 9/x2)dx = dt

So, our equation becomes,

= ∫dt/(t2 + 18)

= (1/3√2) tan-1(t/3√2) +c

= (1/3√2) tan-1[(x2 − 9)/3√2x] + c

### Question 4. ∫1/(x4 + x2 + 1)dx

Solution:

We have,

∫1/(x4 + x2 + 1)dx

= ∫(1/x2)/(x2 + 1 + 1/x2)dx

= (1/2)∫(1 + 1/x2 − 1 + 1/x2)/(x2 + 1 + 1/x2)dx

= (1/2)∫(1 + 1/x2)/[(x − 1/x)2 + 3]dx − (1/2)∫(1 − 1/x2)/(x + 1/x)2− 1)dx

Let x − 1/x = t, so we get, (1 + 1/x2)dx = dt and

Let x + 1/x = y, so we get, (1 − 1/x2)dx = dy.

So, our equation becomes,

= (1/2)∫dt/(t2 + 3) − (1/2)∫dy/(y2 − 1)

= (1/2√3) tan-1(t/√3) − (1/4) log |(y − 1)/(y + 1)| + c

= (1/2√3) tan-1[(x2 − 1)/√3x] − (1/4) log |(x2 + 1 − x)/(x2 + 1 + x)| + c

### Question 5. ∫(x2 − 3x + 1)/(x4 + x2 + 1)dx

Solution:

We have,

∫(x2 − 3x + 1)/(x4 + x2 + 1)dx

= ∫(1 − 3/x + 1/x2)/(x2 + 1 + 1/x2)dx

= ∫(1 + 1/x2)/[(x − 1/x)2 + 3]dx − ∫3x/(x4 + x2 + 1)dx

Let x − 1/x = t, so we get, (1 + 1/x2)dx = dt

Let x2 = y, so we get, 2xdx = dy

So, our equation becomes,

= ∫dt/(t2 + 3) − (3/2)∫dy/(y2 + y + 1)

= ∫dt/(t2 + 3) − (3/2)∫dy/[(y+1/2)2 + 3/4]

= (1/√3) tan-1(t/√3) − (3/2)(2/√3)tan-1[(y + 1/2)/(√3/2)] + c

= (1/√3) tan-1(t/√3) − √3tan-1[(2y + 1)/√3] + c

= (1/√3) tan-1[(x2 − 1)/√3] − √3tan-1[(2x2 + 1)/√3] + c

### Question 6. ∫(x2 + 1)/(x4 − x2 + 1)dx

Solution:

We have,

∫(x2 + 1)/(x4 − x2 + 1)dx

= ∫(1 + 1/x2)/(x2 − 1 + 1/x2)dx

= ∫(1 + 1/x2)/[(x − 1/x)2 + 1]dx

Let x − 1/x = t, so we get, (1 + 1/x2)dx = dt

So, the equation becomes,

= ∫dt/(t2 + 1)

= tan−1t + c

= tan−1[(x2 − 1)/x] + c

### Question 7. ∫(x2 − 1)/(x4 + 1)dx

Solution:

We have,

∫(x2 − 1)/(x4 + 1)dx

= ∫(1 − 1/x2)/(x2 + 1/x2)dx

= ∫(1 − 1/x2)/[(x + 1/x)2 − 2]dx

Let x + 1/x = t, so we get, (1 − 1/x2)dx = dt

So, the equation becomes,

= ∫dt/(t2 − 2)

= (1/2√2) log |(t − √2)/(t + √2)| + c

= (1/2√2) log |(x2 + 1 − √2x)/(x2 + 1 + √2x)| + c

### Question 8. ∫(x2 + 1)/(x4 + 7x2 + 1)dx

Solution:

We have,

∫(x2 + 1)/(x4 + 7x2 + 1)dx

= ∫(1 + 1/x2)/(x2 + 7 + 1/x2)dx

= ∫(1 + 1/x2)/[(x − 1/x)2 + 9]dx

Let x − 1/x = t, so we get, (1 + 1/x2)dx = dt

So, the equation becomes,

= ∫dt/(t2 + 9)

= (1/3)tan−1(t/3) + c

= (1/3)tan−1[(x2 − 1)/3x] + c

### Question 9. ∫(x − 1)2/(x4 + x2 + 1)dx

Solution:

We have,

∫(x − 1)2/(x4 + x2 + 1)dx

= ∫(x2 − 2x + 1)/(x4 + x2 + 1)dx

= ∫(1 − 2/x + 1/x2)/(x2 + 1 + 1/x2)dx

= ∫(1 + 1/x2)/[(x − 1/x)2 + 3]dx − ∫2x/(x4 + x2 + 1)dx

Let x − 1/x = t, so we get, (1 + 1/x2)dx = dt

Let x2 = y, we get, 2xdx = dy

So, our equation becomes,

= ∫dt/(t2 + 3) − ∫dy/(y2 + y + 1)

= ∫dt/(t2 + 3) − ∫dy/[(y + 1/2)2 + 3/4]

= (1/√3) tan−1[(x2 − 1)/√3x] − (2/√3) tan−1[(2x2 + 1)/√3] + c

### Question 10. ∫1/(x4 + 3x2 + 1)dx

Solution:

We have,

∫1/(x4 + 3x2 + 1)dx

= ∫(1/x2)/(x2 + 3 + 1/x2)dx

= (1/2)∫(1 + 1/x2 − 1 + 1/x2)/(x2 + 3 + 1/x2)dx

= (1/2)∫(1 + 1/x2)/[(x − 1/x)2 + 5]dx − (1/2)∫(1 − 1/x2)/[(x + 1/x)2 + 1]dx

Let x − 1/x = t, we get, (1 + 1/x2)dx = dt

Let x + 1/x = y, we get, (1 − 1/x2)dx = dy

So, the equation becomes,

= (1/2)∫dt/(t2 + 5) − (1/2)∫dy/(y2 + 1)

= (1/2√5) tan−1(t/√5) − (1/2) tan−1y + c

= (1/2√5) tan−1[(x2 − 1)/√5x] − (1/2) tan−1[(x2 + 1)/x] + c

### Question 11. ∫1/(sin4x + sin2x cos2x + cos4x)dx

Solution:

We have,

∫1/(sin4x + sin2x cos2x + cos4x)dx

= ∫(1/cos4x)/[(sin4x + sin2x cos2x + cos4x)/(cos4x)]dx

= ∫(sec4x)/(tan4x + tan2x + 1)dx

= ∫[(1 + tan2x)sec2x]/(tan4x + tan2x + 1)dx

Let tan x = t, so we get sec2xdx = dt

So, the equation becomes,

= ∫(1 + t2)/(t4 + t2 + 1)dt

= ∫(1 + 1/t2)/(t2 + 1/t2 + 1)dt

= ∫(1 + 1/t2)/[(t − 1/t)2 + 3]dt

Let t − 1/t = y, so we get, (1 + 1/t2)dt = dy

So, the equation becomes,

= ∫dy/(y2 + 3)

= (1/√3) tan−1(y/√3) + c

= (1/√3) tan−1[(t−1/t)/√3] + c

= (1/√3) tan−1[(tan x − 1/tan x)/√3] + c

= (1/√3) tan−1[(tan x − cot x)/√3] + c

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